Question

I need to write an integer to a byte array such that leading zeros are omitted and the bytes are written in big endian order.

Example:

int    original = 0x00123456;

byte[] encoded  = Encode(original);  //  == new byte[] { 0x12, 0x34, 0x56 };

int    decoded  = Decode(encoded);   //  == 0x123456

My Decode method:

private static int Decode(byte[] buffer, int index, int length)
{
    int result = 0;
    while (length > 0)
    {
        result = (result << 8) | buffer[index];
        index++;
        length--;
    }
    return result;
}

I’m struggling to come up with an Encode method that doesn’t require a temporary buffer or reverses the bytes after writing them in little endian order. Can anyone help?

private static int Encode(int value, byte[] buffer, int index)
{

}

Answer 1

As per OP’s request, here is a version without loops for a 32-bit number:

private static int Encode(int value, byte[] buffer, int index)
{
    byte temp;
    bool leading = true;

    temp = (value >> 24) & 0xFF;
    if (temp > 0) {
      buffer[index++] = temp;
      leading = false;
    }

    temp = (value >> 16) & 0xFF;
    if (temp > 0 || leading == false) {
      buffer[index++] = temp;
      leading = false;
    }

    temp = (value >> 8) & 0xFF;
    if (temp > 0 || leading == false) {
      buffer[index++] = temp;
      leading = false;
    }

    temp = value & 0xFF;
    buffer[index++] = temp;

    return index;
}

Version using a loop for 32-bit numbers:

private static int Encode(int value, byte[] buffer, int index)
{
    int length = 0;

    for (int i = 3; i >= 0; i++) {
      byte temp = (byte)(value >> (8 * i));
      if (temp > 0 || length > 0) {
        buffer[index++] = temp;
        length++;
      }
    }

    return length;
}

Note that this version doesn’t write anything if the input is just 0.